3Sum
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
Solution
Sort the array, and then iterate through the array — for every iteration, have three pointers, where you fix a number, and then select a number smaller and larger than it. For each iteration, you can binary search for a better possibility that matches.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
res = []
n = len(nums)
nums = sorted(nums)
for i in range(n-2):
if i > 0 and nums[i] == nums[i-1]:
continue
j = i+1
k = n-1
new_target = -nums[i]
while j < k:
summ = nums[j] + nums[k]
if summ < new_target:
j += 1
elif summ > new_target:
k -= 1
else:
res.append([nums[i], nums[j], nums[k]])
while j < k and nums[j+1] == nums[j]:
j += 1
j += 1
while k > j and nums[k-1] == nums[k]:
k -= 1
k -= 1
return res